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Question Number 149425 by help last updated on 05/Aug/21

Answered by dumitrel last updated on 05/Aug/21

$$x>0;x\ne 1$$$$\sqrt[4]{{log}_{3}x}=y>0\Rightarrow x=3^{y^4}$$$$9^y-4x^{\frac{1}{y^3}}+3=0\Rightarrow 9^y-4\cdot {\left(3^{y^4}\right)}^{\frac{1}{y^3}}+3=0\Rightarrow$$$${\left(3^y\right)}^2-4\cdot 3^y+3=0\Rightarrow (3^y-1)(3^y-3)=0\Rightarrow$$$$y=1\Rightarrow x=3$$

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